Question

Cos37 Cos53-Sin37 Sin53​

Answer 1

Объяснение:

$cos37^0cos53^0-sin37^0sin53^0=\\=\frac{1}{2}*(cos(37^0-53^0)+cos(37^0+53^0))-\frac{1}{2}*(cos(37^0-53^0)-cos(37^0+53^0))=\\=\frac{1}{2}*(cos(-16^0)+cos90^0-cos(-16^0)+cos90^0)=\frac{1}{2}*2*cos90^0=cos90^0=0.$